A girl of mass 60 kg throws a ball of mass 0.8 kg against a wall. The ball strikes the wall
horizontally with a speed of 11 m/s, and it bounces back with this same speed. The ball is in contact
with the wall 0.05 s. What is the average force exerted on the wall by the ball?
A) 6600 N B) 26,400 N C) 13,200 N D) 180 N E) 350 N
Forcefromwall*time= massball*changeinvelocity.
Notice the change in velocity is 22m/s (Vf-Vi) where V is vectors, with direction.
350 N
To find the average force exerted on the wall by the ball, we can use the impulse-momentum principle.
The impulse experienced by an object is defined as the change in momentum of the object. It can be calculated using the formula:
Impulse = Change in momentum
The change in momentum of the ball can be calculated using the formula:
Change in momentum = Final momentum - Initial momentum
The initial momentum of the ball can be calculated as the product of its mass and initial velocity:
Initial momentum = mass × initial velocity
The final momentum of the ball can be calculated as the product of its mass and final velocity:
Final momentum = mass × final velocity
Since the ball bounces back with the same speed, the final velocity is equal in magnitude but opposite in direction to the initial velocity. Thus, the final velocity is -11 m/s.
Now, let's calculate the initial momentum:
Initial momentum = 0.8 kg × 11 m/s = 8.8 kg·m/s
And the final momentum:
Final momentum = 0.8 kg × (-11 m/s) = -8.8 kg·m/s
The change in momentum is then:
Change in momentum = (-8.8 kg·m/s) - (8.8 kg·m/s) = -17.6 kg·m/s
Since the ball is in contact with the wall for 0.05 s, the impulse experienced by the ball can be calculated using the formula:
Impulse = Average force × Time
Therefore, we have:
-17.6 kg·m/s = Average force × 0.05 s
Rearranging the equation:
Average force = -17.6 kg·m/s ÷ 0.05 s
Average force = -352 N
However, the negative sign indicates that the force is in the opposite direction. To find the magnitude of the force, we take the absolute value:
Average force = |-352 N| = 352 N
Therefore, the average force exerted on the wall by the ball is 352 N.
The correct answer is E) 352 N.
To find the average force exerted on the wall by the ball, we can make use of Newton's second law of motion, which states that force (F) is equal to the rate of change of momentum (dP) with time (dt):
F = dP/dt
The momentum of an object is defined as the product of its mass (m) and velocity (v):
P = mv
In this scenario, we are given the mass of the ball (m = 0.8 kg) and its initial velocity (v = 11 m/s). Since the ball bounces back with the same speed, the final velocity (vf) can be taken as -11 m/s (negative sign indicates opposite direction).
The change in momentum (dP) can be calculated by subtracting the initial momentum (P1) from the final momentum (P2):
dP = P2 - P1
Since the ball is in contact with the wall for a very short period of time (0.05 s), we can use the average force (F) over this time interval.
To calculate the initial momentum (P1):
P1 = m * v1
P1 = 0.8 kg * 11 m/s = 8.8 kg*m/s
To calculate the final momentum (P2):
P2 = m * v2
P2 = 0.8 kg * (-11 m/s) = -8.8 kg*m/s
Now we can calculate the change in momentum (dP):
dP = P2 - P1
dP = (-8.8 kg*m/s) - (8.8 kg*m/s)
dP = -17.6 kg*m/s
Finally, we can calculate the average force (F) using Newton's second law:
F = dP/dt
F = -17.6 kg*m/s / 0.05 s
F = -352 N
Since the force exerted on the wall acts in the opposite direction to the motion of the ball, we take the negative sign into account. So the average force exerted on the wall by the ball is 352 N (approximately).
However, since the question asks for the magnitude of the force, we can take the absolute value. Therefore, the answer is approximately 352 N.
None of the given answer choices match our calculated value, so it seems there may be an error in the problem or answer choices.